前言
在我刚开始做游戏的时候，在学习到三维物体旋转时第一次遇到四元数，那时候只学习到使用欧拉角旋转会遇到旋转死锁的问题，四元数可以解决。自己那时候学习了下四元数，感觉好难理解。最后照着Unity官方文档，只做到了会用，我在学习新东西时候总是想把原理搞个清楚，这样心里才会舒服。这个四元数的问题一直如鲠在喉，最近在看《游戏引擎架构》，对于里面每一个知识点都不想放过去，所以就有了这篇文章。
四元数
复数
要说明白四元数，首先得搞清楚复数的概念，数学家为了解决对负数开平方的问题，引入了复数。
\[z = a + bi\]其中$i^2 = -1$
复数
如果将二维复数扩展到三维，就是四元数
\[q = s + xi + yj + zk\]有序对表达式
\(q = [s, v]\)
v可以单独为组件
\(q = [s, xi + yj + zk]\)
加法
\(\begin{array}{rcl} q_a & = & [s_a,\mathbf{a}] \\ q_b & = & [s_b,\mathbf{b}] \\ q_a+q_b & = & [s_a+s_b,\mathbf{a}+\mathbf{b}] \\ q_a-q_b & = & [s_a-s_b,\mathbf{a}-\mathbf{b}]\end{array}\)
四元数乘法
\(\begin{array}{rcl} q_a & = & [s_a,\mathbf{a}] \\ q_b & = & [s_b,\mathbf{b}] \\ q_aq_b & = & [s_a,\mathbf{a}][s_b,\mathbf{b}] \\ &= &(s_a+ x_ai + y_aj + z_ak)(s_b + x_bi + y_bj + z_bk) \\ &= &(s_as_b - x_ax_b - y_ay_b - z_az_b) \\ & &+(s_ax_b + s_bx_a + y_az_b - y_bz_a)i \\ & &+(s_ay_b + s_by_a + z_ax_b - x_az_b)j\\ & &+(s_az_b + x_ay_b - y_ax_b + s_bz_a)k \end{array}\)
\(i = [0,i] ~~~ j = [0, j] ~~~~ k = [0,k]\) 实数部分用$[1, 0]$
\[\begin{array}{rcl} [s_a,\mathbf{a}][s_b,\mathbf{b}] &= &(s_as_b - x_ax_b - y_ay_b - z_az_b)[1, 0] \\ & &+(s_ax_b + s_bx_a + y_az_b - y_bz_a)[0, i] \\ & &+(s_ay_b + s_by_a + z_ax_b - x_az_b)[0, j]\\ & &+(s_az_b + s_bz_a + x_ay_b - y_ax_b)[0,k] \end{array}\]展开
\[\begin{array}{rcl} [s_a,\mathbf{a}][s_b,\mathbf{b}] &= &[s_as_b - x_ax_b - y_ay_b - z_az_b, 0] \\ & &+[0,(s_ax_b + s_bx_a + y_az_b - y_bz_a)i] \\ & &+[0,(s_ay_b + s_by_a + z_ax_b - x_az_b)j]\\ & &+[0,(s_az_b + s_bz_a + x_ay_b - y_ax_b)k] \end{array}\]虚数部分结合
\[\begin{array}{rcl} [s_a,\mathbf{a}][s_b,\mathbf{b}] &= &[s_as_b - x_ax_b - y_ay_b - z_az_b, 0] \\ &+&[0,s_a(x_bi + y_bj + z_bk) + s_b(x_ai + y_aj + z_ak) \\ &+&(y_az_b - y_bz_a)i + (z_ax_b - x_az_b)j + (x_ay_b - y_ax_b)k]\\ \end{array}\]最终结果
\[\begin{array}{rcl} [s_a,\mathbf{a}][s_b,\mathbf{b}] &= &[s_as_b - x_ax_b - y_ay_b - z_az_b, \\ &&s_a(x_bi + y_bj + z_bk) + s_b(x_ai + y_aj + z_ak) \\ &+&(y_az_b - y_bz_a)i + (z_ax_b - x_az_b)j + (x_ay_b - y_ax_b)k]\\ \end{array}\]当前设定
\[\begin{array}{rcl} \mathbf{a}~ = & x_ai + y_aj + z_ak \\ \mathbf{b}~ = & x_bi + y_bj + z_bk \\ \mathbf{a.b}~ = & x_ax_b + y_ay_b + z_az_b \\ \mathbf{a}\times\mathbf{b}~ = &(y_az_b - y_bz_a)i + (z_ax_b - x_az_b)j + (x_az_b - z_ax_b)k \end{array}\]带入得到结果
\[\begin{array}{rcl} [s_a,\mathbf{a}][s_b,\mathbf{b}] = [s_as_b - \mathbf{a.b}, s_a\mathbf{b} + s_b\mathbf{a} + \mathbf{a}\times\mathbf{b}] \end{array}\]实数四元数
\(\begin{array}{rcl} q & = &[s, 0] \\ q_a & = &[s_a, 0] \\ q_b & = &[s_b, 0] \\ q_aq_b & = &[s_as_b, 0] \end{array}\)
虚数四元数
\(\begin{array}{rcl} q & = &[0, v] \\ q_a & = &[0, v_a] \\ q_b & = &[0, v_b] \\ q_aq_b & = &[-v_av_b, v_a\times\mathbf{v_b}] \end{array}\)
共轭
\(\begin{array}{rcl} q & = &[s, v] \\ q^* & = &[s, -v] \\ qq^* &= &[s^2+v^2, 0] \end{array}\)
归一化
\(q^{\prime} = \frac{q}{\sqrt{s^2 + v^2}}\)
逆四元数
\(qq^{-1} = [1, 0]\) \(q^{-1} = \frac{q^{*}}{|q|^2}\)
点积
\(\begin{array}{rcl} q_1 & = & [s_1, x_1i + y_1j + z_1k] \\ q_2 & = & [s_2, x_2i + y_2j + z_2k] \\ q_1{\cdot}q_2 & = & s_1s_2 + x_1x_2 + y_1y_z + z_1z_2\\ \end{array}\)
余弦
\(cos{\theta} = \frac{q_1.q_2}{|q_1||q_2|}\)
旋转
2d 旋转
\(q = cos{\theta} + sin{\theta}i\)
3d 旋转
\(q = [cos{\theta}, sin{\theta}\mathbf{v}]\)
旋转
矢量p用四元数表示为纯虚数为:
\[p = [0, \mathbf{p}]\]旋转四元数为:
\[q = [s, \lambda\hat{v}]\]乘法
\[\begin{array}{rcl} p^{\prime} &=& qp \\ &=& [s, \lambda\hat{v}][0, \mathbf{p}] \\ &=& [-\lambda\hat{v}\cdot\mathbf{p}, s\cdot\mathbf{p} + \lambda\hat{v}\times\mathbf{p}] \end{array}\]旋转例子一
绕$k$旋转45°
\[\begin{array}{rcl} p &=& [0, 2i] \\ q &=& [\cos{\theta}, \sin{\theta}\hat{v}] \\ \hat{v} &=& [0, k] \\ p^{\prime} &=&\left[\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}k][0,2i\right] \\ &=& \left[0, \sqrt{2}i + \sqrt{2}j\right] \end{array}\]旋转例子二
绕$jk$的中间轴旋转45˚
\[\begin{array}{rcl} p &=& [0, 2i] \\ q &=& [\cos{\theta}, \sin{\theta}\hat{v}] \\ \hat{v} &=& \left[0, \frac{\sqrt{2}}{2}i + \frac{\sqrt{2}}{2}k\right] \\ \end{array}\]计算结果为:
\[\begin{array}{rcl} p^{\prime} &=& \left[-\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}\mathbf{i}+\frac{\sqrt{2}}{2}\mathbf{k}\right)\cdot2\mathbf{i},\frac{\sqrt{2}}{2}2\mathbf{i}+\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}\mathbf{i}+\frac{\sqrt{2}}{2}\mathbf{k}\right)\times2\mathbf{i}\right] \\ &=& [-1, \sqrt{2}\mathbf{i} + j] \end{array}\]矢量表示为纯虚数，旋转结果不是纯虚数，模变为 $\sqrt{3}$
现在来求q的逆矩阵
\[\begin{array}{rcl} q^{-1} &=& [\cos{\theta}, -\sin{\theta}\hat{v}] \\ q^{-1} &=& \left[\frac{\sqrt{2}}{2}, -\frac{1}{2}i -\frac{1}{2}k\right] \end{array}\]乘逆矩阵:
\[\begin{array}{rcl} p^{\prime}q^{-1} &=& \left[-1, \sqrt{2}i + j\right]\left[\frac{\sqrt{2}}{2}, -\frac{1}{2}i -\frac{1}{2}k\right] \\ &=&\frac{1}{2}\left[-\sqrt{2} + \sqrt{2},i+k+2i +\sqrt{2}j -i +\sqrt{2}j+k\right] \\ &=&\frac{1}{2}\left[0, 2k + 2i + 2\sqrt{2}j\right] \\ &=&[0,i +\sqrt{2}j + k] \end{array}\]